Answer:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
Answer:
D. To ensure the cooling process is not affected by surrounding temperature
Explanation:
The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>
The atomic radius increases down a column (group) and decreases along a row
We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
Answer : The concentration of A after 80 min is, 0.100 M
Explanation :
Half-life = 20 min
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = 80 min
a = initial amount of the reactant = 1.6 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the concentration of A after 80 min is, 0.100 M
